How to obtain $y'$ from $e^{y}=x^{\ln x}$?
This is what I did: $$\ln e^y = \ln x^{\ln x}$$ $$y = \ln ^2 x$$ $$y' = \frac{2 \ln x}{x}$$ Is this correct? When I compare it with an online derivative calculator, the result they gave was different, and they used implicit differentiation instead.
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$\begingroup$$$(e^y)'=e^yy'$$ and $$(e^{ln^2x})'=(\ln^2x)'e^{ln^2x}=2\frac{\ln x}xe^{ln^2x}.$$
Then dividing by $e^y=e^{ln^2x}$,
$$y'=2\frac{\ln x}x.$$
As expected, same answer, though a little more tedious.
This is $$g(y)=f(x),\\ g'(y)y'=f'(x),\\ y'=\frac{f'(x)}{g'(y)}=\frac{f'(x)}{g'(g^{-1}(f(x)))}$$ versus $$y=g^{-1}(f(x)),\\ y'=\frac{f'(x)}{g'(g^{-1}(f(x)))} $$ by the formula for the inverse function and the chain rule.
$\endgroup$ $\begingroup$So if we use implicit differentiation:
$$e^{y}\frac{\mathrm{d}y}{\mathrm{d}x}=2\ln(x)x^{\ln(x)-1}$$
But $e^{y} = x^{\ln(x)}$, so:
$$\frac{\mathrm{d}y}{\mathrm{d}x}=2\ln(x) \frac{x^{\ln(x)-1}}{x^{\ln(x)}}=\frac{2\ln(x)}{x}$$
So you are quite right in your answer (and your approach is absolutely fine too!)
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